Introduction
• Calculation of deflections is an important part of
structural analysis
• Excessive beam deflection can be seen as a mode of
failure.
– Extensive glass breakage in tall buildings can be attributed
to excessive deflections
– Large deflections in buildings are unsightly (and unnerving)
and can cause cracks in ceilings and walls.
– Deflections are limited to prevent undesirable vibrations
Beam Deflection
• Bending changes the
initially straight
longitudinal axis of the
beam into a curve that
is called the
Deflection Curve or
Elastic Curve
Beam Deflection
• To determine the deflection curve:
– Draw shear and moment diagram for the beam
– Directly under the moment diagram draw a line for the
beam and label all supports
– At the supports displacement is zero
– Where the moment is negative, the deflection curve is
concave downward.
– Where the moment is positive the deflection curve is
concave upward
– Where the two curve meet is the Inflection Point
Deflected Shape
Example 1
Draw the deflected shape for each of the beams shown
Example 2
Draw the deflected shape for each of the frames shown
Double Integration Method
Elastic‐Beam Theory
• Consider a differential element
of a beam subjected to pure
bending.
• The radius of curvature ρ is
measured from the center of
curvature to the neutral axis
• Since the NA is unstretched,
the dx=ρdθ
Elastic‐Beam Theory
• Applying Hooke’s law and the Flexure formula, we
obtain:
1=M
ρ EI
Elastic‐Beam Theory
• The product EI is referred to as the flexural rigidity.
• Since dx = ρdθ, then
dθ = M dx (Slope)
EI
In most calculus books
d 2v / dx2
1+ (dv / dx)2
[ ]1= 3
2
ρ
d 2v / dx2
1+ (dv / dx)2
[ ]M= 3 (exact solution)
2
EI
d 2v = M
dx2 EI
The Double Integration Method
Relate Moments to Deflections
d 2v = M
dx2 EI
θ (x) = dv = ∫ M(x) dx Do Not
dx Integration Constants
EI(x)
Use Boundary Conditions to
v( x) = ∫∫ M (x) dx 2 Evaluate Integration
EI Constants
(x)
Assumptions and Limitations
Deflections caused by shearing action negligibly small compared to bending
Deflections are small compared to the cross‐sectional dimensions of the beam
All portions of the beam are acting in the elastic range
Beam is straight prior to the application of loads
y L Examples
PL x
M = −PL + Px
x
P P EI d2y = M
@ x dx 2
EI d 2 y = −PL + Px
dx 2
dy x2
Integrating once EI dx = −PLx + P 2 + c1
@ x = 0 dy = 0 ⇒ EI (0) = −PL(0)+ P (0)2 + c1 ⇒ c1 =0
dx
2
EIy = − PLx2 + P x3 +
Integrating twice 2 6(0)3 c2
⇒ EI (0) = − + + c2 ⇒ c2 = 0 PL3
@ x = 0 y = 0 PL (0)2 P 3EI
EIy = − PLx2 + P x3 26
∆ max =
@ x = L y = ymax
2 6 PL L2 L3 PL3 PL3
2 6 6 3EI
EIymax = − + P = − ⇒ ymax = −
y W x M = − W (L − x)2
2
WL2 x
2 WL L EI d2y = M
dx 2
@ x d2y −W (L x)2
EI dx 2 = 2 −
Integrating once EI dy = W (L − x)3 + c1
dx 2
3
@ x = 0 dy = 0 ⇒ EI (0) = W (L − 0)3 + c1 ⇒ c1 = − WL3
dx 6
2 3
∴ EI dy = W (L − x)3 − WL3
dx 6 6
Integrating twice EIy = −W (L − x)4 − WL3 x + c2
6 6
4
@ x = 0 y=0 ⇒ EI (0) = − W (L − 0)4 − WL3 (0) + c2 ⇒ c2 = WL4
6 24
6 4
EIy = − W (L − x)4 − WL3 x + WL4
24 6 24
Max. occurs @ x = L
EIymax = − W L4 + WL4 = − WL4 ⇒ ymax = − WL4
6 24 8 8EI
∆ max = WL4
8EI
Example yx
x
WL L WL
2 M = WL x −Wx x 2
22
EI d2y = WL x −W x2
dx 2 2 2
Integrating EI dy = WL x2 −W x3 + c1
dx 2 2 2 3
Since the beam is symmetric @ x= L dy = 0
⎜⎛ L ⎞3 2 dx
@ x= L ⎛⎜ L ⎞2 −W ⎝2⎠ ⇒ WL3
2 EI (0) = WL ⎝ 2 ⎠ 23 24
+ c1 c1 = −
22
∴ EI dy = WL x2 − W x3 − WL3
dx 4 6 24
Integrating EIy = WL x3 −W x4 − WL3 x + c2
@ x = 0 y = 0 4 3 6 4 24
⇒ EI (0) = WL (0)3 −W (0)4 − WL3 (0)+ c2 ⇒ c2 = 0
4 6 24
3 4
∴ EIy = WL x3 − W x4 − WL3 x
12 24 24
Max. occurs @ x = L /2 EIymax = − 5WL4
384
∆ max = 5WL4
384EI
Example yx P
x
P L/2 L/2 P
2 for 0 < x < L M = P x 2
22
d2y
EI dx 2 = P x for 0 < x < L
2 2
Integrating EI dy = P x2 + c1
dx 2 2
Since the beam is symmetric @ x= L dy = 0
2 dx
⎛⎜ L ⎞2 c1 ⇒
@ x= L EI (0) = P ⎝ 2⎠ + c1 = − PL2
2 2 2 16
∴ EI dy = P x2 − PL2
dx 4 16
Integrating EIy = P x3 − PL2 x + c2
@ x = 0 y = 0 4 3 16
⇒ EI (0) = P (0)3 − PL2 (0) + c2 ⇒ c2 = 0
4 16
3
∴ EIy = P x3 − PL2 x
12 16
Max. occurs @ x = L /2 EIymax = − PL 3
48
∆ max = PL3
48EI
Example
Example 5
Moment‐Area Theorems
Moment‐Area Theorems
Theorem 1: The change in slope between any two points on
the elastic curve equal to the area of the bending moment
diagram between these two points, divided by the product EI.
d 2v = M ⇒ θ = dv
dx 2 EI
dx
dθ = M ⇒ dθ = ⎛ M ⎟⎠⎞ dx
⎝⎜ EI
dx EI
∫θBA B M dx
EI
=
A
dt = xdθ
dθ = ⎛ M ⎞ dx
⎝⎜ EI ⎟⎠
B M dx B M dx
EI EI
∫ ∫tBA= x =x
A A
Moment‐Area Theorems
Theorem 2: The vertical distance of point A on a elastic
curve from the tangent drawn to the curve at B is equal to
the moment of the area under the M/EI diagram between
two points (A and B) about point A .
B∫t A B M dx
EI
=x
A
∫t AB B M dx
EI
=x
A
Example 1
Example 2
Example 3
Example 4
Example 5
Example 6
Example 7
M/EI ‐20/EI
‐30/EI
‐
tC /B = ⎛ − 20 ⋅ 2 ⎞ ⋅ (1) + 1 ⋅ ⎛ − 10 ⋅ 2 ⎞ ⋅ ⎛ 2 ⋅ 2 ⎞
⎜⎝ EI ⎟⎠ 2 ⎝⎜ EI ⎟⎠ ⎝⎜ 3 ⎟⎠
= − 53.33 kN ⋅ m 3 = 0.00741 rad
EI
Another Solution
Conjugate-Beam Method
Conjugate-Beam Method
dV =w d 2M =w
dx dx 2
dθ = M d 2v = M
dx 2 EI
dx EI
Integrating
V = ∫wdx M = ∫ ⎡⎣∫wdx ⎦⎤dx
θ = ∫ ⎛ M ⎞⎠⎟dx v = ∫ ⎡⎢⎣∫ ⎛ M ⎞⎠⎟ dx ⎤
⎝⎜ EI ⎝⎜ EI ⎦⎥dx
Conjugate-Beam Supports
Example 1
Find the Max. deflection Take E=200Gpa, I=60(106)
θB = VB' = − 562.5
EI
∆B = M B' = 562.5 (25) = −14062.5
EI EI
Example 2
Find the deflection at Point C
C